question_answer

Charges \[+q\] and \[-q\] are placed at points A and B respectively which are a distance 2L apart, C is the midpoint between A and B. The work done in moving a charge \[+Q\] from C to D along the semicircle CRD is

A) \[\frac{qQ}{2\pi {{\varepsilon }_{0}}L}\]

B)  \[\frac{qQ}{6\pi {{\varepsilon }_{0}}L}\]

C) \[-\frac{qQ}{6\pi {{\varepsilon }_{0}}L}\]

D)  \[\frac{qQ}{4\pi {{\varepsilon }_{0}}L}\]

Correct Answer: C

Solution :

Potential at \[C={{V}_{C}}=0\]

Potential at \[D={{V}_{D}}\]\[=K\left( \frac{-q}{L} \right)+\frac{Kq}{3L}=-\frac{2}{3}\frac{Kq}{L}\]

Potential difference \[{{V}_{D}}-{{V}_{C}}=\frac{-2}{3}\frac{Kq}{L}=\frac{1}{4\pi {{\in }_{0}}}\left( -\frac{2}{3}.\frac{q}{L} \right)\]

\[\Rightarrow \] Work done \[=Q({{V}_{D}}-{{V}_{C}})\]\[=-\frac{2}{3}\times \frac{1}{4\pi {{\varepsilon }_{0}}}\frac{qQ}{L}=\frac{-qQ}{6\pi {{\varepsilon }_{0}}L}\]